Tombstone Recessions Related to Atmospheric Sulfur Dioxide Concentrations

Carbonate Rocks

Limestone belongs to a class of rock type called carbonates. Carbonates are derived from organic matter. All carbonates are highly soluble, hence why many caverns are formed in limestone rock. Under immense heat and pressure, rocks undergo metamorphism. This changes the crystal properties of the rock and makes it more dense and compacted. Marble is the metamorphosed form of limestone. It is harder and denser but it still remains soluble, especially in the presence of acids. Nowadays, the ease with which carbonates dissolve is well known. With this knowledge, we have transitioned to more durable stone such as granite, schist, etc., to mark our deceased.

Marble Rock Face (Wikipedia)

\(SO_2\) Weathering

Sulphur dioxide (\(SO_2\)) is a chemical compound that is released into the atmosphere from the burning of coal. This compound conjoins with meteroic water to form acid rain, playing a big part in tombstone dissolution. However, from Meierdings’ study, it is surmised that the direct reaction of \(SO_2\) with the calcite (\(CaCO_3\)) minerals within the marble is responsible for most of the weathering. This reaction drives the production of \(CaSO_4\), and in the presence of water vapor it produces \(CaSO_4-H_2O\), a sulphate mineral called gypsum. Gypsum takes up 38% more volume than \(CaCO_3\) so that when it crystallizes in the pore space of the statue it induces granular breakage and loss. For this reason, regions with higher \(SO_2\) atmospheric concentrations will experience higher weathering rates in carbonate materials.

Tombstone Recession (Meierding, 1993)

My Interest in the data

As a geology major, my interest in geologic reasons behind various processes is higher than most. The relationship between carbonates and dissolution is a very large geologic topic. This creates a wide range of problems outside of tombstones including sinkhole formation, road degradation, \(CaCO_3\) buildup inside of pipes, etc. Also, tombstones mark something of immense value, whether the people underneath of them are remembered by anyone or not. They signify a life lived, however long or brief, and whatever we can do to ensure some memory of them lasts seems worthwhile to me.

The problem to be solved

We want to discover the relationship of \(SO_2\) concentration vs. the average recession rates of the marble tombstones within those regions. If we see a linear relationship we may surmise that recession is closely intertwined with \(SO_2\) emmisions.

Viewing this preliminary data we can see that there does seem to be a linear type relationship beteen \(SO_2\) concentrations and recession rates over 100 year periods. A further look at the mathematical relationship will validate this possibility.

Estimators of \(\beta_0\) and \(\beta_1\)

## (Intercept)          so 
## 0.322995899 0.008593333

Using this information we can see that \[\hat{\beta}_0=0.322996\] \[\hat{\beta}_1=0.008593\]

95% Confidence Interval of Estimates

##             95 % C.I.lower    95 % C.I.upper
## (Intercept)        0.00445           0.64155
## so                 0.00661           0.01058

Our confidence interval for \(\hat{\beta}_0\) is [0.00445, 0.64155]. Our initial estimate is 0.322996 so we can see that the estimate lies in this 95% confidence interval.

Least Squares Estimate

Using our intitial estimate equation \(\hat{y}=\hat{\beta}_0+\hat{\beta}_1x_i\) with our estimators we get: \[\hat{y}=\hat{\beta}_0 + \hat{\beta}_1x_i=0.322996 + 0.008593x_i\]. This shows us that for every +1 increase in \(SO_2\) mean concentration over a 100 year period, we get an increase of 0.008593 mm of recession. Or more simply stated, for every +100 increase of mean \(SO_2\) concentration over a 100 year period we get an increase of 0.8593 mm of recession.

Normality of Error Distribution

I have used a Shapiro-Wilk test to test for normality. It can be observed that the error distribution appears to be normal. When we look at the p-value of the error distribution it is 0.828. This very high p-value is much greater than our alpha value of 0.05. With a null hypothesis of normal distribution, we will accept this due to the high p-value. Therefore we will say that error is distributed normally.\[\epsilon \sim N(0,\sigma^2)\]

Residual Data

Using a residual vs fitted plot we can see a uniformity about 0 of the residuals. This information tells us that there is no non-uniform and significant deviation from the fitted line from the model.

Here we can see how many outlier residuals we actually have.

This graph gives us a visual representation of the data residuals about the fitted line. We are able to see specific datapoints in which there are high residual distances. This information gives us the RSS (residual sum of squares), the predicted deviations from the fitted line.

## [1] 2.53143

Plot of Means

From this graph we can visually see that we are graphing the difference between the mean \(SO_2\) concentration and the fitted line at specific data points. This will provide us with the model sum of squares (MSS).

## [1] 10.90307

Means with Total Devaitions

From this graph we can see a visual representaion of total deviation from the mean recession amount. This gives us an idea about distance from the mean line in relation to mean \(SO_2\) concentrations. This provides us with the total sum of squares (TSS).

## [1] 13.4345

\(R^2\) Calculation

\(R^2\) is calcualted with MSS/TSS. This gives us an idea of how well the data conforms to the fitted line of the model. A, \(R^2\) value of 1 means that all of the data falls perfectly onto the fitted line. An \(R^2\) value close to one means that the data follows a well developed linear trend.

## [1] 0.8115724

Our \(R^2\) value is 0.8116, showing a well developed linear trend.

Summary of Dataset without 18th Data Point

## 
## Call:
## lm(formula = ree ~ soo, data = tb2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -0.5750 -0.1613 -0.0046  0.1045  0.6607 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 0.2883429  0.1377363   2.093   0.0507 .  
## soo         0.0091315  0.0008851  10.317 5.52e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3284 on 18 degrees of freedom
## Multiple R-squared:  0.8553, Adjusted R-squared:  0.8473 
## F-statistic: 106.4 on 1 and 18 DF,  p-value: 5.516e-09

Summary of Original Dataset

## 
## Call:
## lm(formula = re ~ so)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.72384 -0.19138  0.06136  0.13320  0.69412 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 0.3229959  0.1521958   2.122   0.0472 *  
## so          0.0085933  0.0009499   9.046 2.58e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.365 on 19 degrees of freedom
## Multiple R-squared:  0.8116, Adjusted R-squared:  0.8017 
## F-statistic: 81.83 on 1 and 19 DF,  p-value: 2.579e-08

## (Intercept)          so 
## 0.322995899 0.008593333

Comparing these two datasets we can see that with the large outlier removed our Multiple \(R^2\) values rises from 0.8116 to 0.8553 and our residual standard error is reduced from 0.365 to 0.3284. This indicates that the model without the large outlier may be a better fit to the data. However, the adjusted \(R^2\) value is lower in the first model. This suggests that the first model may actually be more indicative of the true regression than the new model.

Model Predictions

The creation of these models gives us a very powerful tool to predict likely conditions in the future. We will predict what the amount of recession will be in tombstones over a 100 year period with mean \(SO_2\) concentrations at 150, 250, and 500 (ug/m^3).

##        1        2        3 
## 1.611996 2.471329 4.619662

With \(SO_2\) concentrations of 50, 250, and 500 (ug/m^3) per 100 years we should see 1.612, 2.471, and 4.620 mm recessions, respectively.